Expression of the Equations of 3D Egg Shape in terms of degrees

Question

I'd basically like to have 3D version of this article section or this section. So for my case, there are two angles for latitude and longitude to construct 3D egg. Any hint to extend the formula to 3D would be appreciated.

Answer 1

So, you want the parametric form of the surface of revolution of a chosen curve.

If your curve given by some functions $x = f(\theta)$ and $y = g(\theta)$, the surface of revolution about the $x$-axis is parametrized by $$\begin{align} x &= f(\theta), \\ y &= g(\theta)\cos(\phi), \\ z &= g(\theta)\sin(\phi), \end{align}$$ where $\theta$ has the same range in as your input curve and $\phi$ goes from $0$ to $360$ degrees.

Answer 2

The link to the article that was originally cited for this question has gone stale, so I thought I would post the egg shape that I came up with back in 2004: $$ \begin{align} x(\phi)&=\left\{\begin{array}{} \cos(\phi)&\text{if }-\pi\le\phi\lt0\\ -1+2\cos(\phi)&\text{if }0\le\phi\lt\frac\pi4\\ \left(2-\sqrt2\right)\cos(\phi)\phantom{1+\,}&\text{if }\frac\pi4\le\phi\lt\frac{3\pi}4\\ 1+2\cos(\phi)&\text{if }\frac{3\pi}4\le\phi\lt\pi \end{array}\right.\\ y(\phi)&=\left\{\begin{array}{} \sin(\phi)&\text{if }-\pi\le\phi\lt0\\ 2\sin(\phi)&\text{if }0\le\phi\lt\frac\pi4\\ 1+\left(2-\sqrt2\right)\sin(\phi)&\text{if }\frac\pi4\lt\phi\lt\frac{3\pi}4\\ 2\sin(\phi)&\text{if }\frac{3\pi}4\le\phi\lt\pi\\ \end{array}\right. \end{align} $$

This was originally visualized as arcs of four circles:

but it can also be seen as the envelope of a family of circles:

In $3$ dimensions, we can look at the surface of revolution:

The $3$ dimensional egg is also an envelope of a family of spheres:

The parametric equation for the $3$ dimensional egg is $$ \bbox[5px,border:2px solid #C0A000]{\begin{align} x(\phi,\theta)&=\left\{\begin{array}{} \cos(\phi)\cos(\theta)&\text{if }-\frac\pi2\le\phi\lt0\\ (-1+2\cos(\phi))\cos(\theta)&\text{if }0\le\phi\lt\frac\pi4\\ \left(2-\sqrt2\right)\cos(\phi)\cos(\theta)&\text{if }\frac\pi4\le\phi\le\frac\pi2 \end{array}\right.\\ y(\phi,\theta)&=\left\{\begin{array}{} \cos(\phi)\sin(\theta)&\text{if }-\frac\pi2\le\phi\lt0\\ (-1+2\cos(\phi))\sin(\theta)&\text{if }0\le\phi\lt\frac\pi4\\ \left(2-\sqrt2\right)\cos(\phi)\sin(\theta)\,&\text{if }\frac\pi4\le\phi\le\frac\pi2 \end{array}\right.\\ z(\phi,\theta)&=\left\{\begin{array}{} \sin(\phi)&\text{if }-\frac\pi2\le\phi\lt0\\ 2\sin(\phi)&\text{if }0\le\phi\lt\frac\pi4\\ 1+\left(2-\sqrt2\right)\sin(\phi)\phantom{(|\,)}&\text{if }\frac\pi4\le\phi\le\frac\pi2 \end{array}\right. \end{align}} $$